Question

A juggler maintains four balls in motion making each of them to rise a height of 20 m from his hand. What time interval should be maintained for the proper distance between them?

• A. \(1.5\,\text{s}\)
• B. \(\dfrac{3}{2}\,\text{s}\)
• C. \(1\,\text{s}\)
• D. \(2\,\text{s}\)

Hint:

For vertical motion: \[ h = \frac{u^2}{2g}, \quad T = \frac{2u}{g} \] If multiple objects are kept in the air simultaneously, divide the total time of flight by the number of objects to get the time interval.

Answer 1

The Correct Option is C

Step 1: Each ball is projected vertically upwards and reaches a maximum height of \(h = 20\,\text{m}\). Using the relation: \[ h = \frac{u^2}{2g} \] \[ 20 = \frac{u^2}{2 \times 10} \] \[ u^2 = 400 \Rightarrow u = 20\,\text{m/s} \]
Step 2: Time of flight for one ball: \[ T = \frac{2u}{g} = \frac{2 \times 20}{10} = 4\,\text{s} \]
Step 3: Since the juggler keeps four balls simultaneously in motion, they must be thrown at equal time intervals. \[ \text{Time interval} = \frac{T}{4} = \frac{4}{4} = 1\,\text{s} \]